Question

A bit (meaning binary digit) is 0 or 1. An ordered array of eight bits (such as01101001) is a byte. How many different bytes are there? If you select a byte at random, what is the probability that you select 11000010? What is the probability thatyou select a byte containing three 1’s and five 0’s?

Short answer

Answer

The total different arrays that can be formed is $2^B$, the probability of selecting 11000010 is$2^{-B}$, probability of selecting a byte containing three 1’s and five 0’s is $\frac{7}{32}$.

Step-by-step solution

Given Information

An ordered array of 8 bits is provided

Definition of Independent Event

When the order of arrangement is definite, the permutation is applied and when the order is not definite,combination is applied.

Finding the number of bytes that can be created.

There are 8 digits that are to be filled with 0 or 1, this implies that the total different arrays that can be formed is$2^B$.

Finding the probability of selecting 11000010

There are 8 digits that are to be filled with 0 or 1, this implies that each place has an equal probability of getting filled by 0 or 1 and each digit is independent of other, this implies that probability of selecting 11000010 is$2^{-B}$.

Finding the probability ofselecting a byte containing three 1’s and five 0’s

There are three 1s and five 0s and thus the possible arrangements can be $\frac{8!}{3!5!}$.

This implies that the number of outcomes favorable are $\frac{8!}{3!5!}$and total number of outcomes are $2^B$.

Apply the formula for probability, that is $p=\frac{numberofoutcomesfavorabletoE}{totalnumberofoutcomes}$to get the probability of selecting a byte containing three 1’s and five 0’s.

Hence the desired probability is$\frac{7}{32}$.